Question

A frictionless inclined plane is 6 m long and rests on a wall that is 2 m high.

How much force is needed to push a block of ice weighing 300 N up the plane?
A) 600 NB) 200 NC) 300 ND) 50 NE) 100 N

Answer 1

The force needed to push a block of ice is 100 N.

Step-by-step explanation:

It is given that,

Length of the friction less plane, l = 6 m

Height of the wall, h = 2 m

Weight of the block of ice, W = 300 N

The length of the inclined plane and the height of the wall act as hypotenuse and the perpendicular of the right angled triangle. Using trigonometry to find the angle as :

`sin\theta=(P)/(H)`

`sin\theta=(2)/(6)`

`sin\theta=(1)/(3)`

Force needed to push a block of ice weighing 300 N up the plane is equal to the horizontal component of the force as :

`F_x=F\ sin\theta`

`F_x=300* (1)/(3)`

`F_x=100\ N`

So, the force needed to push a block of ice is 100 N. Hence, this is the required solution.